Question

Решите задания 3, 4, 5, 6 на листке

Answer 1

Ответ:

3) $\frac{27m^3-n^3}{4n^2-m^2} : \frac{9m^2+3mn+n^2}{m-2n} =\frac{(3m-n)*(9m^2+3mn+n^2)}{(2n-m)*(2n+m)} * \frac{m-2n}{9m^2+3mn+n^2} =\frac{3m-n}{(2n-m)*(2n+m)} *$

$*(-(2n-m)) =\frac{3m-n}{2n+m} * (-1)=-\frac{3m-n}{2n+m} *1=\frac{n-3m}{2n+m}$

4) $\frac{(a-b)*(a+b)-(a-b)^2}{ab-b^2} =\frac{a^2-b^2-(a-b)^2}{b*(a-b)} =\frac{(a-b)*(a+b)-(a-b)^2}{b*(a-b)} =\frac{(a-b)*(a+b-(a-b))}{b*(a-b)} =$

$=\frac{a+b-a+b}{b} = \frac{2b}{b} =2$

5) $\frac{2-5x-2y+5xy}{10x^3-9x^2+2x} =\frac{-(-2+5x)+y*(-2+5x)}{x*(10x^2-9x+2)} =\frac{-(-2+5x)*(1-y)}{x*(10x^2-4x-5x+2)} = \frac{-(-2+5x)*(1-y)}{x*(2x*(5x-2)-(5x-2))} =$

$= \frac{-(-2=5x)*(1-y)}{x*(5x-2)*(2x-1)} =\frac{-(1-y)}{x*(2x-1)} =-\frac{1-y}{x*(2x-1)} =\frac{y-1}{x*(2x-1)} = \frac{y-1}{2x^2-x}$

6) $9x^2-4y^2+4y-1=9x^2-(4y^2-4y+1)=9x^2-(2y-1)^2=(3x-(2y-1))*$

$*(3x+(2y-1))= (3x-2y+1)*(3x+2y-1)$